Two tanks of similar volume are full of a mixture of oil and water. In

Tank #1: oil/water = 5/8, volume V
Tank #2: oil/water =7/19, volume V

Oil per combined colume
= (5/13*V + 7/26*V)/(V + V)
= (10+7)/26/2
= 17/52

Water per combined volume is 1-17/52 = 35/52

Oil:water=17:35

FINAL ANSWER IS (C)

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Two tanks of similar volume are full of a mixture of oil and water. In the first, the ratio of oil and water is 5:8 and in the second, it is 7:19. If both these tanks are poured in a larger tank, what would be the resultant ratio of oil and water?

A. 17:52
B. 1:3
C. 17:35
D. 151:304
E. 1:2

Tanker 1: \(oil:water = 5x:8x\)
\(Vol_A\) = 13x

Tanker 2: \(oil:water = 7y:19y\)
\(Vol_B\) = 26x

\(Vol_A\) = \(Vol_B\); x = 2y

Larger Tanker: \(oil:water\) = 5x+7y: 8x+19y = 10y+7y: 16y+19y = 17:35
Ans: C

Assume total volume as 26 litres .

Oil : water = 17/35

The answer is C

Let us consider there are 26 units of each tank

Thus in Tank A , Ratio of Oil:Water becomes 10:16
and in tank B , Ratio of Oil:Water becomes 7:19

Therefore adding together , the ratio is 17:35

First tank = 5:8 = 10:16 --> 10 liters oil and 16 liters water, for a total of 26 liters
Second tank = 7:19 --> 7 liters oil and 19 liters water, for a total of 26 liters
When the tanks are combined, we get:
\(\frac{oil}{water} = \frac{10+7}{16+19} = \frac{17}{35}\)

Two tanks of similar volume are full of a mixture of oil and water. In the first, the ratio of oil and water is 5:8 and in the second, it is 7:19. If both these tanks are poured in a larger tank, what would be the resultant ratio of oil and water?

A. 17:52
B. 1:3
C. 17:35
D. 151:304
E. 1:2

as both are equal vol, 5x+8x=7y+19y
13x=26y
x=2y.
ratio, 5x+7y/8x+19y => 17y/35y
so,17/35
Ans C

since volumes are equal for both containers

thus volume of oil will be
=\(\frac{5}{3 }+\frac{ 7}{26} \)/ 2

=\( \frac{17}{52}\)
or oil and water ratio will be
17 : (52-17)
= 17:35

C IMO

Let the volume of tank 1 be KX and tank 2 be KY
KX=KY(Given)
First tank contain 5x:8x=13x(total) (oil:water)
Second tank contain 7y:19y=26y (oil:water)
13x should be equal to 26y
13x*2=26y
26x=26y
Now. Since the ratio is doubled in first tank.
The new ratio will be 10x:16x
Pouring both tank solution in a larger rank
10xy+7xy:16xy+19xy
17xy:35xy
17:35

OA should be C

Posted from my mobile device

Two tanks of similar volume are full of a mixture of oil and water. In the first, the ratio of oil and water is 5:8 and in the second, it is 7:19. If both these tanks are poured in a larger tank, what would be the resultant ratio of oil and water?

A. 17:52
B. 1:3
C. 17:35
D. 151:304
E. 1:2

Let's check the given options. Here Oil to Water of final mixture are given. We an see that Oil component is 17 for two of the options and 1 for other two.
BUT Water component in a given options are different. So, if we sole for Water component we can solve the question at the earliest.

Tank One Oil to Water ratio = 5/8 So,
Water to total Mixture = \(\frac{8}{13}\)

Tank Two Oil to Water ratio = \(\frac{7}{19}\)
Water to Total mixture = \(\frac{19}{26}\)

Water in new mixture = \(\frac{8}{13} + \frac{19}{26} = \frac{35}{26}\) (At this point we can stop and mark C as our answer)

Otherwise
Oil to total mixture ratio = \(\frac{5}{13}\) in first tank AND Oil to total mixture ratio = \(\frac{7}{26}\) in second tank
Oil in new mixture = \(\frac{5}{13} + \frac{7}{26} = \frac{17}{26}\)

Oil to Water ratio in new mixture = \(\frac{17}{26} / \frac{35}{26} = \frac{17}{35}\)

Answer C.

Let the volume of one tank be 'x'. Since both tanks have similar volumes, they both have volume as 'x'.
Let the proportion of Oil in new mixture be 'y' and Water be 'z'.
therefore,

(5/13)x + (7/26)x = y(2x)
=> 17/52 = y
=> 1-17/52 = z = 35/52
=> y:z = 17:35

Ans:C

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